Because the generating sets are disjoint, the <free product> has the presentation
$$
\boxed{G*H=\langle X\sqcup Y\mid R\sqcup S\rangle.}
$$
This is immediate from the <universal property of a group presentation>: a map out of this presented group is exactly a pair of homomorphisms from $G$ and $H$ to the target group.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 104 / 1 / a / Solution

A map from the generators $S$ of $\langle S\mid R\rangle$ to a group $G$ extends to a homomorphism exactly when every relator in $R$ evaluates to the identity. The extension is then unique.


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 Incoming links: Universal property of a group presentation